Hyperbola
Hyperbola Standard Form / Area
nta_pyq_2023_jan
Grade 11
Question:
Let T and C respectively be the transverse and conjugate axes of the hyperbola $16x^2 - y^2 + 64x + 4y + 44 = 0$. Then the area of the region above the parabola $x^2 = y + 4$, below the transverse axis T and on the right of the conjugate axis C is:
4\sqrt{6} + \dfrac{44}{3}
4\sqrt{6} + \dfrac{28}{3}
4\sqrt{6} - \dfrac{44}{3}
4\sqrt{6} - \dfrac{28}{3}
Step-by-Step Solution
Key Concept: Rewrite hyperbola in standard form by completing the square; find transverse and conjugate axes, then integrate.
$16(x+2)^2-(y-2)^2=16$, center $(-2,2)$. Transverse axis: $y=2$, conjugate axis: $x=-2$. Area between $x^2=y+4$ and $y=2$ for $x \ge -2$: $4\sqrt{6}-\frac{28}{3}$.
Correct Answer: 4